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Mar 24, 2021 at 13:59 comment added Iosif Pinelis @Strickland : (i) Such multiple roots cannot exist, because $R(u)u^{l-k}$ has a nonzero derivative in those right and left neighborhoods. I have now added this additional explanation. (ii) I have also added details on $m(p)<m(q)$.
Mar 24, 2021 at 13:56 history edited Iosif Pinelis CC BY-SA 4.0
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Mar 24, 2021 at 6:41 comment added Strickland Thanks for the clarification. If still have some questions though, if you do not mind. 1. When you argue that $R(u)u^{l-k}+s=0$ has no more than two roots $u$ resp. one, I assume you base this on monotonicity. However, could it not have a root of multiplicity 4 for example? 2. It is immediately assumed that $m(p)<m(q)$. What if they are equal?
Mar 24, 2021 at 6:35 vote accept Strickland
Mar 23, 2021 at 16:23 history edited Iosif Pinelis CC BY-SA 4.0
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Mar 23, 2021 at 14:20 history edited Iosif Pinelis CC BY-SA 4.0
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Mar 23, 2021 at 14:11 comment added Iosif Pinelis @Strickland : Thank you for your comment. Somehow, I thought only of the roots of $q$ in $(0,1)$ of odd multiplicities. This is now fixed.
Mar 23, 2021 at 14:05 history edited Iosif Pinelis CC BY-SA 4.0
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Mar 23, 2021 at 13:58 history edited Iosif Pinelis CC BY-SA 4.0
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Mar 23, 2021 at 6:41 comment added Strickland Thanks a lot. If I understand correctly, the argument for case 1 seems a continuity argument of sort $\lim_{t_n\to t_*} m(q_{t_n})=m(q_{t_*})>m_p=0$, which leads to the contradiction. Did I understand this correctly? If so, is it clear that $m()$ is continuous and secondly, why wouldn't this hold for case 2 as well?
Mar 22, 2021 at 22:02 history edited Iosif Pinelis CC BY-SA 4.0
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Mar 22, 2021 at 19:23 history undeleted Iosif Pinelis
Mar 22, 2021 at 17:50 history deleted Iosif Pinelis via Vote
Mar 22, 2021 at 17:39 history answered Iosif Pinelis CC BY-SA 4.0